Advertisements
Advertisements
प्रश्न
In Fig below we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.
उत्तर
We have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm
In ΔECD and ΔEAB
∠CED = ∠AEB [common]
∠ECD = ∠EAB [corresponding angles]
Then, ΔECD ~ ΔEAB ….(i) [By AA similarity]
`therefore"EC"/"EA"="CD"/"AB"` [Corresponding parts of similar Δ are proportional]
`rArr"EC"/"EA"=x/6` .........(ii)
In ΔACD and ΔAEF
∠CAD = ∠EAF [common]
∠ACD = ∠AEF [corresponding angles]
Then, ΔACD ~ ΔAEF [By AA similarity]
`therefore"AC"/"AE"="CD"/"EF"`
`rArr"AC"/"AE"=x/10` ..........(iii)
Add equations (iii) & (ii)
`therefore"EC"/"EA"+"AC"/"AE"=x/6+x/10`
`rArr"AE"/"AE"=(5x+3x)/30`
`rArr1=(8x)/30`
`rArrx=30/8=3.75` cm
From (i) "DC"/"AB"="ED"/"BE"
`rArr3.75/6=y/(y+4)`
⇒ 6y = 3.75y + 15
⇒ 2.25y = 15
`rArry=15/2.25=6.67` cm
APPEARS IN
संबंधित प्रश्न
In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
ΔPDC ∼ ΔBEC
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that `("AB")/("PQ") = ("AD")/("PM")`.
In the following figure, PA, QB and RC are each perpendicular to AC. Prove that `1/x+1/z=1/y`
The sides of certain triangles are given below. Determine which of them right triangles are.
1.6cm, 3.8cm, 4cm
The sides of certain triangles are given below. Determine which of them right triangles are.
(a – 1) cm, `2sqrta` cm, (a + 1) cm
A ladder 10m long reaches the window of a house 8m above the ground. Find the distance of the foot of the ladder from the base of the wall.
The corresponding sides of two similar triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.
In the given figure MN|| BC and AM: MB= 1: 2
find ` (area(ΔAMN))/(area(ΔABC))`
In figure, if DE || BC, find the ratio of ar(ADE) and ar(DECB).
If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
